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Question Number 27564 by Tinkutara last updated on 09/Jan/18

Commented by Tinkutara last updated on 09/Jan/18

Commented by mrW2 last updated on 12/Jan/18

Commented by mrW2 last updated on 12/Jan/18

$$\frac{1}{2}{mv}^2=mgR(1-\sin \theta )$$$$\Rightarrow v^2=2gR(1-\sin \theta )$$$$mg\sin \theta -N=\frac{{mv}^2}{R}=2mg(1-\sin \theta )$$$$\Rightarrow N=3mg(\sin \theta -\frac{2}{3})$$$$with{\theta }_1={\sin }^{-1}\frac{2}{3}\approx 42°$$$$for90°>\theta >42°:N>0$$$$for\theta <42°:N<0$$$$$$$$\Rightarrow Answer4iscorrect.$$

Commented by Tinkutara last updated on 12/Jan/18

Thank you very much Sir! I got the answer.

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